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Homework Help: ODE question

  1. Nov 9, 2012 #1
    I want to solve:
    [itex] y'=\frac{(1+y)^2} {x(y+1)-x^2} [/itex]

    What I tried:
    I have no basis to think that y' is positive or negative in some domain, but if I do, I can write:
    [itex] x'(y)=\frac {x(y+1)-x^2}{(1+y)^2}=\frac{x}{(1+y)} +\frac{x^2}{(1+y)^2}[/itex]

    and then I can substitute [itex]z=\frac{x}{(1+y)}[/itex]

    And I get the following ODE: [itex](y+1)\frac{dz}{dy}+z=z-z^2[/itex].

    So the solution is [itex]\frac{1}{z}=\frac{(y+1)}{x}=\ln|y+1|+C [/itex].

    Then I can mark all this as "draft" and write: Lets notice that if [itex]F(x,y)=\frac{(y+1)}{x}-\ln|y+1|=C [/itex] is a potential, so this is indeed the solution.

    Is this solution legit?
    How can I solve this problem alternatively?

    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2


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    You might want to check that. I get something a little simpler.
    Sorry, I don't follow your logic there. Did you check that your solution satisfies the original equation?
    Try substituting z = (y+1)/x straight off.
  4. Nov 9, 2012 #3
    This is what I've got: attached pdf document.
    Can you help me to spot what I did wrong?

    Attached Files:

    • 1.b.pdf
      File size:
      231.9 KB
  5. Nov 9, 2012 #4


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    Hmm I only looked at this for a moment, but it looks like a POSSIBLE candidate for the method of successive approximations.
  6. Nov 9, 2012 #5
    This is a Bernoulli ODE with n = 2, with respect to [itex]x(y)[/itex] and is transformed to a linear ODE with the substitution:
    z = x^{1 - n} = x^{-1}
    The derivative is:
    x' = -x^{-2} \, x'
    So, multiply the equation by [itex](-x^{-2})[/itex], and see what you get.
  7. Nov 9, 2012 #6

    But, can I assume that y' is positive or negative in some domain?
    How can I explain this?
  8. Nov 9, 2012 #7
    [itex]x'[/itex] and [itex]y'[/itex] have the same sign.
  9. Nov 9, 2012 #8
    I understand this. but how I can understand that any of one of the tho preserve sign in any domain?
  10. Nov 9, 2012 #9
    Ok, so:
    y' = \frac{(1 + y)^2}{x (y + 1) - x^2} = \frac{(1 + y)^2}{x (y + 1 - x)}
    The sign of [itex]y'[/itex] is the same as the sign of [itex]x (y + 1 - x)[/itex]. You may draw the regions in the x-y plane where this is positive, negative, or zero. When it is zero, [itex]y'[/itex] diverges, but [itex]x' = 0[/itex] (unless [itex]y = -1[/itex]).

    When you find the general solution, try and see if any particular solution crosses from one region to another.
  11. Nov 9, 2012 #10
    Thank you very much!
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